3.28.62 \(\int \frac {-16 x^2+8 x^3+8 x^4+e^2 (2+8 x+8 x^2)+e (-9-4 x-16 x^2-16 x^3)}{2 x^2+8 x^3+8 x^4+e^2 (2+8 x+8 x^2)+e (-4 x-16 x^2-16 x^3)} \, dx\)

Optimal. Leaf size=27 \[ 3+\frac {3}{2 \left (2+\frac {1}{x}\right ) \left (-\frac {e}{3}+\frac {x}{3}\right )}+x \]

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Rubi [A]  time = 0.14, antiderivative size = 23, normalized size of antiderivative = 0.85, number of steps used = 4, number of rules used = 4, integrand size = 96, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {1680, 1814, 21, 8} \begin {gather*} x-\frac {9 x}{2 \left (-2 x^2+(2 e-1) x+e\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16*x^2 + 8*x^3 + 8*x^4 + E^2*(2 + 8*x + 8*x^2) + E*(-9 - 4*x - 16*x^2 - 16*x^3))/(2*x^2 + 8*x^3 + 8*x^4
+ E^2*(2 + 8*x + 8*x^2) + E*(-4*x - 16*x^2 - 16*x^3)),x]

[Out]

x - (9*x)/(2*(E + (-1 + 2*E)*x - 2*x^2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {-(1+2 e)^2 \left (35-4 e-4 e^2\right )+288 (1-2 e) x-32 \left (19+4 e+4 e^2\right ) x^2+256 x^4}{\left (1+4 e+4 e^2-16 x^2\right )^2} \, dx,x,\frac {1}{32} (8-16 e)+x\right )\\ &=-\frac {9 x}{2 \left (e+(-1+2 e) x-2 x^2\right )}-\frac {\operatorname {Subst}\left (\int \frac {-2 (1+2 e)^4+32 (1+2 e)^2 x^2}{1+4 e+4 e^2-16 x^2} \, dx,x,\frac {1}{32} (8-16 e)+x\right )}{2 (1+2 e)^2}\\ &=-\frac {9 x}{2 \left (e+(-1+2 e) x-2 x^2\right )}+\operatorname {Subst}\left (\int 1 \, dx,x,\frac {1}{32} (8-16 e)+x\right )\\ &=x-\frac {9 x}{2 \left (e+(-1+2 e) x-2 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 52, normalized size = 1.93 \begin {gather*} \frac {1}{2} \left (-\frac {9}{(1+2 e) (-1-2 e+2 (e-x))}-\frac {9 e}{(1+2 e) (e-x)}-2 (e-x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*x^2 + 8*x^3 + 8*x^4 + E^2*(2 + 8*x + 8*x^2) + E*(-9 - 4*x - 16*x^2 - 16*x^3))/(2*x^2 + 8*x^3 +
8*x^4 + E^2*(2 + 8*x + 8*x^2) + E*(-4*x - 16*x^2 - 16*x^3)),x]

[Out]

(-9/((1 + 2*E)*(-1 - 2*E + 2*(E - x))) - (9*E)/((1 + 2*E)*(E - x)) - 2*(E - x))/2

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fricas [B]  time = 0.53, size = 45, normalized size = 1.67 \begin {gather*} \frac {4 \, x^{3} + 2 \, x^{2} - 2 \, {\left (2 \, x^{2} + x\right )} e + 9 \, x}{2 \, {\left (2 \, x^{2} - {\left (2 \, x + 1\right )} e + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+8*x+2)*exp(1)^2+(-16*x^3-16*x^2-4*x-9)*exp(1)+8*x^4+8*x^3-16*x^2)/((8*x^2+8*x+2)*exp(1)^2+(-
16*x^3-16*x^2-4*x)*exp(1)+8*x^4+8*x^3+2*x^2),x, algorithm="fricas")

[Out]

1/2*(4*x^3 + 2*x^2 - 2*(2*x^2 + x)*e + 9*x)/(2*x^2 - (2*x + 1)*e + x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+8*x+2)*exp(1)^2+(-16*x^3-16*x^2-4*x-9)*exp(1)+8*x^4+8*x^3-16*x^2)/((8*x^2+8*x+2)*exp(1)^2+(-
16*x^3-16*x^2-4*x)*exp(1)+8*x^4+8*x^3+2*x^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/2*(8*sageVARx/4+(72*exp(2)*exp(1)+36*e
xp(2)+72*exp(1)^2+54*exp(1)+9)/(4*exp(2)+4*exp(1)+1)^2/(2*sageVARx+1)+(-36*exp(2)+36*exp(1)^2)/(16*exp(2)^2+32
*exp(2)*exp(1)+8*exp(

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maple [A]  time = 0.10, size = 25, normalized size = 0.93

method	result	size
risch	\(x -\frac {9 x}{4 \left (x \,{\mathrm e}-x^{2}+\frac {{\mathrm e}}{2}-\frac {x}{2}\right )}\)	\(25\)
norman	\(\frac {\left (2 \,{\mathrm e}^{2}-{\mathrm e}-4\right ) x -2 x^{3}+{\mathrm e}^{2}-\frac {{\mathrm e}}{2}}{\left (2 x +1\right ) \left ({\mathrm e}-x \right )}\)	\(45\)
gosper	\(\frac {4 \,{\mathrm e}^{2} x -4 x^{3}+2 \,{\mathrm e}^{2}-2 x \,{\mathrm e}-{\mathrm e}-8 x}{4 x \,{\mathrm e}-4 x^{2}+2 \,{\mathrm e}-2 x}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^2+8*x+2)*exp(1)^2+(-16*x^3-16*x^2-4*x-9)*exp(1)+8*x^4+8*x^3-16*x^2)/((8*x^2+8*x+2)*exp(1)^2+(-16*x^3
-16*x^2-4*x)*exp(1)+8*x^4+8*x^3+2*x^2),x,method=_RETURNVERBOSE)

[Out]

x-9/4*x/(x*exp(1)-x^2+1/2*exp(1)-1/2*x)

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maxima [A]  time = 0.34, size = 26, normalized size = 0.96 \begin {gather*} x + \frac {9 \, x}{2 \, {\left (2 \, x^{2} - x {\left (2 \, e - 1\right )} - e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+8*x+2)*exp(1)^2+(-16*x^3-16*x^2-4*x-9)*exp(1)+8*x^4+8*x^3-16*x^2)/((8*x^2+8*x+2)*exp(1)^2+(-
16*x^3-16*x^2-4*x)*exp(1)+8*x^4+8*x^3+2*x^2),x, algorithm="maxima")

[Out]

x + 9/2*x/(2*x^2 - x*(2*e - 1) - e)

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mupad [B]  time = 1.86, size = 20, normalized size = 0.74 \begin {gather*} x+\frac {9\,x}{2\,\left (2\,x+1\right )\,\left (x-\mathrm {e}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2)*(8*x + 8*x^2 + 2) - exp(1)*(4*x + 16*x^2 + 16*x^3 + 9) - 16*x^2 + 8*x^3 + 8*x^4)/(exp(2)*(8*x + 8*
x^2 + 2) - exp(1)*(4*x + 16*x^2 + 16*x^3) + 2*x^2 + 8*x^3 + 8*x^4),x)

[Out]

x + (9*x)/(2*(2*x + 1)*(x - exp(1)))

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sympy [A]  time = 0.46, size = 22, normalized size = 0.81 \begin {gather*} x + \frac {9 x}{4 x^{2} + x \left (2 - 4 e\right ) - 2 e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**2+8*x+2)*exp(1)**2+(-16*x**3-16*x**2-4*x-9)*exp(1)+8*x**4+8*x**3-16*x**2)/((8*x**2+8*x+2)*exp
(1)**2+(-16*x**3-16*x**2-4*x)*exp(1)+8*x**4+8*x**3+2*x**2),x)

[Out]

x + 9*x/(4*x**2 + x*(2 - 4*E) - 2*E)

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